5.11: The F Distribution (2024)

  1. Last updated
  2. Save as PDF
  • Page ID
    10351
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vectorC}[1]{\textbf{#1}}\)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    \(\newcommand{\R}{\mathbb{R}}\) \(\newcommand{\N}{\mathbb{N}}\) \(\newcommand{\E}{\mathbb{E}}\) \(\newcommand{\P}{\mathbb{P}}\) \(\newcommand{\var}{\text{var}}\) \(\newcommand{\sd}{\text{sd}}\) \(\newcommand{\skw}{\text{skew}}\) \(\newcommand{\kur}{\text{kurt}}\)

    In this section we will study a distribution that has special importance in statistics. In particular, this distribution arises from ratios of sums of squares when sampling from a normal distribution, and so is important in estimation and in the two-sample normal model and in hypothesis testing in the two-sample normal model.

    Basic Theory

    Definition

    Suppose that \(U\) has the chi-square distribution with \(n \in (0, \infty)\) degrees of freedom, \(V\) has the chi-square distribution with \(d \in (0, \infty)\) degrees of freedom, and that \(U\) and \(V\) are independent. The distribution of \[ X = \frac{U / n}{V / d} \] is the \(F\) distribution with \(n\) degrees of freedom in the numerator and \(d\) degrees of freedom in the denominator.

    The \(F\) distribution was first derived by George Snedecor, and is named in honor of Sir Ronald Fisher. In practice, the parameters \( n \) and \( d \) are usually positive integers, but this is not a mathematical requirement.

    Distribution Functions

    Suppose that \(X\) has the \( F \) distribution with \( n \in (0, \infty) \) degrees of freedom in the numerator and \( d \in (0, \infty) \) degrees of freedom in the denominator. Then \( X \) has a continuous distribution on \( (0, \infty) \) with probability density function \( f \) given by \[ f(x) = \frac{\Gamma(n/2 + d/2)}{\Gamma(n / 2) \Gamma(d / 2)} \frac{n}{d} \frac{[(n/d) x]^{n/2 - 1}}{\left[1 + (n / d) x\right]^{n/2 + d/2}}, \quad x \in (0, \infty) \] where \( \Gamma \) is the gamma function.

    Proof

    The trick, once again, is conditioning. The conditional distribution of \( X \) given \( V = v \in (0, \infty) \) is gamma with shape parameter \( n/2 \) and scale parameter \( 2 d / n v \). Hence the conditional PDF is \[ x \mapsto \frac{1}{\Gamma(n/2) \left(2 d / n v\right)^{n/2}} x^{n/2 - 1} e^{-x(nv /2d)} \] By definition, \( V \) has the chi-square distribution with \( d \) degrees of freedom, and so has PDF \[ v \mapsto \frac{1}{\Gamma(d/2) 2^{d/2}} v^{d/2 - 1} e^{-v/2} \] The joint PDF of \( (X, V) \) is the product of these functions: \[g(x, v) = \frac{1}{\Gamma(n/2) \Gamma(d/2) 2^{(n+d)/2}} \left(\frac{n}{d}\right)^{n/2} x^{n/2 - 1} v^{(n+d)/2 - 1} e^{-v( n x / d + 1)/2}; \quad x, \, v \in (0, \infty)\] The PDF of \( X \) is therefore \[ f(x) = \int_0^\infty g(x, v) \, dv = \frac{1}{\Gamma(n/2) \Gamma(d/2) 2^{(n+d)/2}} \left(\frac{n}{d}\right)^{n/2} x^{n/2 - 1} \int_0^\infty v^{(n+d)/2 - 1} e^{-v( n x / d + 1)/2} \, dv \] Except for the normalizing constant, the integrand in the last integral is the gamma PDF with shape parameter \( (n + d)/2 \) and scale parameter \( 2 d \big/ (n x + d) \). Hence the integral evaluates to \[ \Gamma\left(\frac{n + d}{2}\right) \left(\frac{2 d}{n x + d}\right)^{(n + d)/2} \] Simplifying gives the result.

    Recall that the beta function \( B \) can be written in terms of the gamma function by \[ B(a, b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a + b)},\ \quad a, \, b \in (0, \infty) \] Hence the probability density function of the \( F \) distribution above can also be written as \[ f(x) = \frac{1}{B(n/2, d/2)} \frac{n}{d} \frac{[(n/d) x]^{n/2 - 1}}{\left[1 + (n / d) x\right]^{n/2 + d/2}}, \quad x \in (0, \infty) \] When \( n \ge 2 \), the probability density function is defined at \( x = 0 \), so the support interval is \( [0, \infty) \) is this case.

    In the special distribution simulator, select the \(F\) distribution. Vary the parameters with the scroll bars and note the shape of the probability density function. For selected values of the parameters, run the simulation 1000 times and compare the empirical density function to the probability density function.

    Both parameters influence the shape of the \( F \) probability density function, but some of the basic qualitative features depend only on the numerator degrees of freedom. For the remainder of this discussion, let \( f \) denote the \( F \) probability density function with \( n \in (0, \infty) \) degrees of freedom in the numerator and \( d \in (0, \infty) \) degrees of freedom in the denominator.

    Probability density function \( f \) satisfies the following properties:

    1. If \( 0 \lt n \lt 2 \), \( f \) is decreasing with \( f(x) \to \infty \) as \( x \downarrow 0 \).
    2. If \( n = 2 \), \( f \) is decreasing with mode at \( x = 0 \).
    3. If \( n \gt 2 \), \(f\) increases and then decreases, with mode at \(x = \frac{(n - 2) d}{n (d + 2)}\).
    Proof

    These properties follow from standard calculus. The first derivative of \( f \) is \[ f^\prime(x) = \frac{1}{B(n/2, d/2)} \left(\frac{n}{d}\right)^2 \frac{[(n/d)x]^{n/2-2}}{[1 + (n/2)x]^{n/2 + d/2 + 1}} [(n/2 - 1) - (n/d)(d/2 + 1)x], \quad x \in (0, \infty) \]

    Qualitatively, the second order properties of \( f \) also depend only on \( n \), with transitions at \( n = 2 \) and \( n = 4 \).

    For \( n \gt 2 \), define \begin{align} x_1 & = \frac{d}{n} \frac{(n - 2)(d + 4) - \sqrt{2 (n - 2)(d + 4)(n + d)}}{(d + 2)(d + 4)} \\ x_2 & = \frac{d}{n} \frac{(n - 2)(d + 4) + \sqrt{2 (n - 2)(d + 4)(n + d)}}{(d + 1)(d + 4)} \end{align} The probability density function \( f \) satisfies the following properties:

    1. If \( 0 \lt n \le 2 \), \( f \) is concave upward.
    2. If \( 2 \lt n \le 4 \), \( f \) is concave downward and then upward, with inflection point at \( x_2 \).
    3. If \( n \gt 4 \), \( f \) is concave upward, then downward, then upward again, with inflection points at \( x_1 \) and \( x_2 \).
    Proof

    These results follow from standard calculus. The second derivative of \( f \) is \[ f^{\prime\prime}(x) = \frac{1}{B(n/2, d/2)} \left(\frac{n}{d}\right)^3 \frac{[(n/d)x]^{n/2-3}}{[1 + (n/d)x]^{n/2 + d/2 + 2}}\left[(n/2 - 1)(n/2 - 2) - 2 (n/2 - 1)(d/2 + 2) (n/d) x + (d/2 + 1)(d/2 + 2)(n/d)^2 x^2\right], \quad x \in (0, \infty) \]

    The distribution function and the quantile function do not have simple, closed-form representations. Approximate values of these functions can be obtained from the special distribution calculator and from most mathematical and statistical software packages.

    In the special distribution calculator, select the \(F\) distribution. Vary the parameters and note the shape of the probability density function and the distribution function. In each of the following cases, find the median, the first and third quartiles, and the interquartile range.

    1. \(n = 5\), \(d = 5\)
    2. \(n = 5\), \(d = 10\)
    3. \(n = 10\), \(d = 5\)
    4. \(n = 10\), \(d = 10\)

    The general probability density function of the \( F \) distribution is a bit complicated, but it simplifies in a couple of special cases.

    Special cases.

    1. If \( n = 2 \), \[ f(x) = \frac{1}{(1 + 2 x / d)^{1 + d / 2}}, \quad x \in (0, \infty) \]
    2. If \( n = d \in (0, \infty)\), \[ f(x) = \frac{\Gamma(n)}{\Gamma^2(n/2)} \frac{x^{n/2-1}}{(1 + x)^n}, \quad x \in (0, \infty)\]
    3. If \( n = d = 2 \), \[ f(x) = \frac{1}{(1 + x)^2}, \quad x \in (0, \infty) \]
    4. If \( n = d = 1 \), \[ f(x) = \frac{1}{\pi \sqrt{x}(1 + x)}, \quad x \in (0, \infty) \]

    Moments

    The random variable representation in the definition, along with the moments of the chi-square distribution can be used to find the mean, variance, and other moments of the \( F \) distribution. For the remainder of this discussion, suppose that \(X\) has the \(F\) distribution with \(n \in (0, \infty)\) degrees of freedom in the numerator and \(d \in (0, \infty)\) degrees of freedom in the denominator.

    Mean

    1. \(\E(X) = \infty\) if \(0 \lt d \le 2\)
    2. \(\E(X) = \frac{d}{d - 2}\) if \(d \gt 2\)
    Proof

    By independence, \( \E(X) = \frac{d}{n} \E(U) \E\left(V^{-1}\right) \). Recall that \( \E(U) = n \). Similarly if \( d \le 2 \), \( \E\left(V^{-1}\right) = \infty \) while if \( d \gt 2 \), \[ \E\left(V^{-1}\right) = \frac{\Gamma(d/2 - 1)}{2 \Gamma(d/2)} = \frac{1}{d - 2} \]

    Thus, the mean depends only on the degrees of freedom in the denominator.

    Variance

    1. \(\var(X)\) is undefined if \(0 \lt d \le 2\)
    2. \(\var(X) = \infty\) if \(2 \lt d \le 4\)
    3. If \(d \gt 4\) then \[ \var(X) = 2 \left(\frac{d}{d - 2} \right)^2 \frac{n + d - 2}{n (d - 4)} \]
    Proof

    By independence, \( \E\left(X^2\right) = \frac{d^2}{n^2} \E\left(U^2\right) \E\left(V^{-2}\right) \). Recall that \[ E(\left(U^2\right) = 4 \frac{\Gamma(n/2 + 2)}{\Gamma(n/2)} = (n + 2) n \] Similarly if \( d \le 4 \), \( \E\left(V^{-2}\right) = \infty \) while if \( d \gt 4 \), \[ \E\left(V^{-2}\right) = \frac{\Gamma(d/2 - 2)}{4 \Gamma(d/2)} = \frac{1}{(d - 2)(d - 4)} \] Hence \( \E\left(X^2\right) = \infty \) if \( d \le 4 \) while if \( d \gt 4 \), \[ \E\left(X^2\right) = \frac{(n + 2) d^2}{n (d - 2)(d - 4)} \] The results now follow from the previous result on the mean and the computational formula \( \var(X) = \E\left(X^2\right) - \left[\E(X)\right]^2 \).

    In the simulation of the special distribution simulator, select the \(F\) distribution. Vary the parameters with the scroll bar and note the size and location of the mean \( \pm \) standard deviation bar. For selected values of the parameters, run the simulation 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation..

    General moments. For \( k \gt 0 \),

    1. \(\E\left(X^k\right) = \infty\) if \(0 \lt d \le 2 k\)
    2. If \(d \gt 2 k\) then \[ \E\left(X^k\right) = \left( \frac{d}{n} \right)^k \frac{\Gamma(n/2 + k) \, \Gamma(d/2 - k)}{\Gamma(n/2) \Gamma(d/2)} \]
    Proof

    By independence, \( \E\left(X^k\right) = \left(\frac{d}{n}\right)^k \E\left(U^k\right) \E\left(V^{-k}\right) \). Recall that \[ \E\left(U^k\right) = \frac{2^k \Gamma(n/2 + k)}{\Gamma(n/2)} \] On the other hand, \( \E\left(V^{-k}\right) = \infty \) if \( d/2 \le k \) while if \( d/2 \gt k \), \[ \E\left(V^{-k}\right) = \frac{2^{-k} \Gamma(d/2 - k)}{\Gamma(d/2)} \]

    If \( k \in \N \), then using the fundamental identity of the gamma distribution and some algebra, \[ \E\left(X^{k}\right) = \left(\frac{d}{n}\right)^k \frac{n (n + 2) \cdots [n + 2(k - 1)]}{(d - 2)(d - 4) \cdots (d - 2k)} \] From the general moment formula, we can compute the skewness and kurtosis of the \( F \) distribution.

    Skewness and kurtosis

    1. If \( d \gt 6 \), \[ \skw(X) = \frac{(2 n + d - 2) \sqrt{8 (d - 4)}}{(d - 6) \sqrt{n (n + d - 2)}} \]
    2. If \( d \gt 8 \), \[ \kur(X) = 3 + 12 \frac{n (5 d - 22)(n + d - 2) + (d - 4)(d-2)^2}{n(d - 6)(d - 8)(n + d - 2)} \]
    Proof

    These results follow from the formulas for \( \E\left(X^k\right) \) for \( k \in \{1, 2, 3, 4\} \) and the standard computational formulas for skewness and kurtosis.

    Not surprisingly, the \( F \) distribution is positively skewed. Recall that the excess kurtosis is \[ \kur(X) - 3 = 12 \frac{n (5 d - 22)(n + d - 2) + (d - 4)(d-2)^2}{n(d - 6)(d - 8)(n + d - 2)}\]

    In the simulation of the special distribution simulator, select the \(F\) distribution. Vary the parameters with the scroll bar and note the shape of the probability density function in light of the previous results on skewness and kurtosis. For selected values of the parameters, run the simulation 1000 times and compare the empirical density function to the probability density function.

    Relations

    The most important relationship is the one in the definition, between the \( F \) distribution and the chi-square distribution. In addition, the \( F \) distribution is related to several other special distributions.

    Suppose that \(X\) has the \(F\) distribution with \(n \in (0, \infty)\) degrees of freedom in the numerator and \(d \in (0, \infty)\) degrees of freedom in the denominator. Then \(1 / X\) has the \(F\) distribution with \(d\) degrees of freedom in the numerator and \(n\) degrees of freedom in the denominator.

    Proof

    This follows easily from the random variable interpretation in the definition. We can write \[ X = \frac{U/n}{V/d} \] where \( U \) and \( V \) are independent and have chi-square distributions with \( n \) and \( d \) degrees of freedom, respectively. Hence \[ \frac{1}{X} = \frac{V/d}{U/n} \]

    Suppose that \(T\) has the \(t\) distribution with \(n \in (0, \infty)\) degrees of freedom. Then \(X = T^2\) has the \(F\) distribution with 1 degree of freedom in the numerator and \(n\) degrees of freedom in the denominator.

    Proof

    This follows easily from the random variable representations of the \( t \) and \( F \) distributions. We can write \[ T = \frac{Z}{\sqrt{V/n}} \] where \( Z \) has the standard normal distribution, \( V \) has the chi-square distribution with \( n \) degrees of freedom, and \( Z \) and \( V \) are independent. Hence \[ T^2 = \frac{Z^2}{V/n} \] Recall that \( Z^2 \) has the chi-square distribution with 1 degree of freedom.

    Our next relationship is between the \( F \) distribution and the exponential distribution.

    Suppose that \( X \) and \( Y \) are independent random variables, each with the exponential distribution with rate parameter \( r \in (0, \infty) \). Then \(Z = X / Y\). has the \( F \) distribution with \( 2 \) degrees of freedom in both the numerator and denominator.

    Proof

    We first find the distribution function \( F \) of \( Z \) by conditioning on \( X \): \[ F(z) = \P(Z \le z) = \P(Y \ge X / z) = \E\left[\P(Y \ge X / z \mid X)\right] \] But \( \P(Y \ge y) = e^{-r y} \) for \( y \ge 0 \) so \( F(z) = \E\left(e^{-r X / z}\right) \). Also, \( X \) has PDF \( g(x) = r e^{-r x} \) for \( x \ge 0 \) so \[ F(z) = \int_0^\infty e^{- r x / z} r e^{-r x} \, dx = \int_0^\infty r e^{-r x (1 + 1/z)} \, dx = \frac{1}{1 + 1/z} = \frac{z}{1 + z}, \quad z \in (0, \infty) \] Differentiating gives the PDF of \( Z \) \[ f(z) = \frac{1}{(1 + z)^2}, \quad z \in (0, \infty) \] which we recognize as the PDF of the \( F \) distribution with 2 degrees of freedom in the numerator and the denominator.

    A simple transformation can change a variable with the \( F \) distribution into a variable with the beta distribution, and conversely.

    Connections between the \( F \) distribution and the beta distribution.

    1. If \( X \) has the \( F \) distribution with \( n \in (0, \infty) \) degrees of freedom in the numerator and \( d \in (0, \infty) \) degrees of freedom in the denominator, then \[ Y = \frac{(n/d) X}{1 + (n/d) X} \] has the beta distribution with left parameter \( n/2 \) and right parameter \( d/2 \).
    2. If \( Y \) has the beta distribution with left parameter \( a \in (0, \infty) \) and right parameter \( b \in (0, \infty) \) then \[ X = \frac{b Y}{a(1 - Y)} \] has the \( F \) distribution with \( 2 a \) degrees of freedom in the numerator and \( 2 b \) degrees of freedom in the denominator.
    Proof

    The two statements are equivalent and follow from the standard change of variables formula. The function \[ y = \frac{(n/d) x}{1 + (n/d) x} \] maps \( (0, \infty) \) one-to-one onto (0, 1), with inverse \[ x = \frac{d}{n}\frac{y}{1 - y} \] Let \( f \) denote the PDF of the \( F \) distribution with \( n \) degrees of freedom in the numerator and \( d \) degrees of freedom in the denominator, and let \( g \) denote the PDF of the beta distribution with left parameter \( n/2 \) and right parameter \( d/2 \). Then \( f \) and \( g \) are related by

    1. \( g(y) = f(x) \frac{dx}{dy} \)
    2. \( f(x) = g(y) \frac{dy}{dx} \)

    The \( F \) distribution is closely related to the beta prime distribution by a simple scale transformation.

    Connections with the beta prime distributions.

    1. If \( X \) has the \( F \) distribution with \( n \in (0, \infty) \) degrees of freedom in the numerator and \( d \in (0, \infty) \) degrees of freedom in the denominator, then \( Y = \frac{n}{d} X \) has the beta prime distribution with parameters \( n/2 \) and \( d/2 \).
    2. If \( Y \) has the beta prime distribution with parameters \( a \in (0, \infty) \) and \( b \in (0, \infty) \) then \( X = \frac{b}{a} X \) has the \( F \) distribution with \( 2 a \) degrees of the freedom in the numerator and \( 2 b \) degrees of freedom in the denominator.
    Proof

    Let \( f \) denote the PDF of \( X \) and \( g \) the PDF of \( Y \).

    1. By the change of variables formula, \[ g(y) = \frac{d}{n} f\left(\frac{d}{n} y\right), \quad y \in (0, \infty) \] Substituting into the beta \( F \) PDF shows that \( Y \) has the appropriate beta prime distribution.
    2. Again using the change of variables formula, \[ f(x) = \frac{a}{b} g\left(\frac{a}{b} x\right), \quad x \in (0, \infty) \] Substituting into the beta prime PDF shows that \( X \) has the appropriate \( F \) PDF.

    The Non-Central \( F \) Distribution

    The \( F \) distribution can be generalized in a natural way by replacing the ordinary chi-square variable in the numerator in the definition above with a variable having a non-central chi-square distribution. This generalization is important in analysis of variance.

    Suppose that \(U\) has the non-central chi-square distribution with \(n \in (0, \infty) \) degrees of freedom and non-centrality parameter \(\lambda \in [0, \infty)\), \(V\) has the chi-square distribution with \(d \in (0, \infty)\) degrees of freedom, and that \(U\) and \(V\) are independent. The distribution of \[ X = \frac{U / n}{V / d} \] is the non-central \(F\) distribution with \(n\) degrees of freedom in the numerator, \(d\) degrees of freedom in the denominator, and non-centrality parameter \( \lambda \).

    One of the most interesting and important results for the non-central chi-square distribution is that it is a Poisson mixture of ordinary chi-square distributions. This leads to a similar result for the non-central \( F \) distribution.

    Suppose that \( N \) has the Poisson distribution with parameter \( \lambda / 2 \), and that the conditional distribution of \( X \) given \( N \) is the \( F \) distribution with \( N + 2 n \) degrees of freedom in the numerator and \( d \) degrees of freedom in the denominator, where \( \lambda \in [0, \infty) \) and \( n, \, d \in (0, \infty) \). Then \( X \) has the non-central \( F \) distribution with \( n \) degrees of freedom in the numerator, \( d \) degrees of freedom in the denominator, and non-centrality parameter \( \lambda \).

    Proof

    As in the theorem, let \( N \) have the Poisson distribution with parameter \( \lambda / 2 \), and suppose also that the conditional distribution of \( U \) given \( N \) is chi-square with \( n + 2 N \) degrees of freedom, and that \( V \) has the chi-square distribution with \( d \) degrees of freedom and is independent of \( (N, U) \). Let \( X = (U / n) \big/ (V / d) \). Since \( V \) is independent of \( (N, U) \), the variable \( X \) satisfies the condition in the theorem; that is, the conditional distribution of \( X \) given \( N \) is the \( F \) distribution with \( n + 2 N \) degrees of freedom in the numerator and \( d \) degrees of freedom in the denominator. But then also, (unconditionally) \( U \) has the non-central chi-square distribution with \( n \) degrees of freedom in the numerator and non-centrality parameter \( \lambda \), \( V \) has the chi-square distribution with \( d \) degrees of freedom, and \( U \) and \( V \) are independent. So by definition \( X \) has the \( F \) distribution with \( n \) degrees of freedom in the numerator, \( d \) degrees of freedom in the denominator, and non-centrality parameter \( \lambda \).

    From the last result, we can express the probability density function and distribution function of the non-central \( F \) distribution as a series in terms of ordinary \( F \) density and distribution functions. To set up the notation, for \( j, k \in (0, \infty) \) let \( f_{j k} \) be the probability density function and \( F_{j k} \) the distribution function of the \( F \) distribution with \( j \) degrees of freedom in the numerator and \( k \) degrees of freedom in the denominator. For the rest of this discussion, \( \lambda \in [0, \infty) \) and \( n, \, d \in (0, \infty) \) as usual.

    The probability density function \( g \) of the non-central \( F \) distribution with \( n \) degrees of freedom in the numerator, \( d \) degrees of freedom in the denominator, and non-centrality parameter \( \lambda \) is given by \[ g(x) = \sum_{k = 0}^\infty e^{-\lambda / 2} \frac{(\lambda / 2)^k}{k!} f_{n + 2 k, d}(x), \quad x \in (0, \infty) \]

    The distribution function \( G \) of the non-central \( F \) distribution with \( n \) degrees of freedom in the numerator, \( d \) degrees of freedom in the denominator, and non-centrality parameter \( \lambda \) is given by \[ G(x) = \sum_{k = 0}^\infty e^{-\lambda / 2} \frac{(\lambda / 2)^k}{k!} F_{n + 2 k, d}(x), \quad x \in (0, \infty) \]

    5.11: The F Distribution (2024)

    FAQs

    What is the 5th percentile of the F distribution? ›

    The upper fifth percentile is the F-value x such that the probability to the right of x is 0.05, and therefore the probability to the left of x is 0.95.

    How do you find the F distribution value? ›

    The F-distribution results from the quotient of two chi-square distributions which are divided by the respective degrees of freedom. Here you can either calculate the critical F-value or the p-value with given degrees of freedom. You can also read the critical F-value for a given alpha level in the tables below.

    What is the limit of the F distribution? ›

    The F distribution is an asymmetric distribution that has a minimum value of 0, but no maximum value. The curve reaches a peak not far to the right of 0, and then gradually approaches the horizontal axis the larger the F value is. The F distribution approaches, but never quite touches the horizontal axis.

    Can F distribution be greater than 1? ›

    Since variances are always positive, if the null hypothesis is false, MSbetween will generally be larger than MSwithin. Then the F-ratio will be larger than one. However, if the population effect is small, it is not unlikely that MSwithin will be larger in a given sample.

    Is it good to be in the 5th percentile? ›

    There is no one ideal number. Healthy children come in all shapes and sizes, and a baby who is in the 5th percentile can be just as healthy as a baby who is in the 95th percentile.

    What is the significance level of the F test at 5%? ›

    To test the supplier quality example at the 5% level, the F statistic (5.897) is compared to the F table critical value (somewhere between 3.316 and 3.150). Since the F statistic is larger, the result is significant: There are significant differences among your suppliers in terms of average quality level (p < 0.05).

    How to interpret F value? ›

    The F ratio is the ratio of two mean square values. If the null hypothesis is true, you expect F to have a value close to 1.0 most of the time. A large F ratio means that the variation among group means is more than you'd expect to see by chance.

    Is the F distribution always positive? ›

    The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom.

    How to calculate mean of F distribution? ›

    The F distribution has the following properties:
    1. The mean of the distribution is equal to v2 / ( v2 - 2 ) for v2 > 2.
    2. The variance is equal to [ 2 * v22 * ( v1 + v1 - 2 ) ] / [ v1 * ( v2 - 2 )2 * ( v2 - 4 ) ] for v2 > 4.

    What does F-distribution tell us? ›

    The F-distribution was developed by Fisher to study the behavior of two variances from random samples taken from two independent normal populations. In applied problems we may be interested in knowing whether the population variances are equal or not, based on the response of the random samples.

    What is the decision rule for the F-distribution? ›

    Decision Rule: Determine the critical value of F from the F-distribution tables for a chosen significance level (α, often 0.05) and the degrees of freedom from both samples. If the calculated F-statistic is greater than the critical value from the F-distribution table, reject H0.

    What is the lowest value you can get in an F-distribution? ›

    It has a minimum value of zero; there is no maximum value. The distribution's peak happens just to the right of zero and the higher the f-value after that point, the lower the curve. The F distribution is actually a collection of distribution curves.

    What is a significant F-ratio? ›

    A larger calculated F-ratio means that the between-group differences were statistically significant so we can reject the null hypothesis. While an F-ratio smaller than the F-value obtained from a table indicates the groups are too similar so we must accept the null hypothesis.

    What does the F-test tell you? ›

    F test is a statistical test that is used in hypothesis testing to check whether the variances of two populations or two samples are equal or not. In an f test, the data follows an f distribution. This test uses the f statistic to compare two variances by dividing them.

    Can an F-value be less than 1? ›

    All of these F-values would also be associated with fairly large p-values. When F is less than one, we would not reject the hypothesis of no differences.

    What is the 5th percentile in statistics? ›

    To find the 5th percentile for Z (or the cutoff point where 5% of the population lies below it), look at the Z-table and find the probability that's closest to 0.05. You see that the closest probability to 0.05 is either 0.0495 or 0.0505 (use 0.0505 in this case).

    What z-score is the 5th percentile? ›

    For the 5th percentile, we would use Z=-1.645.

    What is the T value of 5th percentile? ›

    t-percentile0.900.975
    21.894.30
    31.643.18
    41.532.78
    51.482.57
    34 more rows

    What is the 5 quantile of the normal distribution? ›

    When we look into PDF, the 5th quantile is the point that cuts off an area of 5% in the lower tail of the distribution: The lower 5% quantile for normal distribution N(0,1). Image by the author. The area under PDF on the left from the red line is exactly 5% of the total area under the curve.

    Top Articles
    Latest Posts
    Article information

    Author: Wyatt Volkman LLD

    Last Updated:

    Views: 6117

    Rating: 4.6 / 5 (46 voted)

    Reviews: 93% of readers found this page helpful

    Author information

    Name: Wyatt Volkman LLD

    Birthday: 1992-02-16

    Address: Suite 851 78549 Lubowitz Well, Wardside, TX 98080-8615

    Phone: +67618977178100

    Job: Manufacturing Director

    Hobby: Running, Mountaineering, Inline skating, Writing, Baton twirling, Computer programming, Stone skipping

    Introduction: My name is Wyatt Volkman LLD, I am a handsome, rich, comfortable, lively, zealous, graceful, gifted person who loves writing and wants to share my knowledge and understanding with you.